LeetCode – Arrays – Last Stone Weight

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

• If x == y, both stones are totally destroyed;
• If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

class Solution {
    public int lastStoneWeight(int[] stones) {
        List<Integer> stone_list = Arrays.stream(stones).boxed().collect(Collectors.toList());
        Collections.sort(stone_list);
        int size = stone_list.size();
        while(true){
            if(size == 0) return 0;
            else if(size == 1) return stone_list.get(0);
            else{
                int x = stone_list.get(size-1);
                int y = stone_list.get(size-2);
                stone_list.remove(size-1);
                stone_list.remove(size-2);
                if(x != y){
                    stone_list.add(x-y);
                }
                Collections.sort(stone_list);
                size = stone_list.size();
            }
        }
    }
}